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Chemical Thermodynamics Formulas For JEE 2027
Chemical Thermodynamics in JEE Chemistry deals with energy changes during chemical reactions and physical processes. The chapter covers heat, work, internal energy, enthalpy, entropy, and Gibbs free energy, along with the laws of thermodynamics. These concepts help students calculate heat changes, assess spontaneity under specified conditions, and understand the relationship between Gibbs free energy and chemical equilibrium.
To prepare effectively, focus on sign conventions, consistent units, and the conditions required for each equation. Practise problems involving calorimetry, Hess’s law, enthalpy changes, and Gibbs free energy. Use a JEE Mains formula sheet PDF to revise essential thermodynamic equations, then apply them in numerical questions. Taking a JEE Mains mock test can help you identify gaps in understanding, review calculation errors, and improve time management.
First Law of Thermodynamics
Types of Systems
| Type | Energy Exchange? | Matter Exchange? |
|---|---|---|
| Open System | Yes | Yes |
| Closed System | Yes | No |
| Isolated System | No | No |
First Law of Thermodynamics
$$$\Delta U = q + w$$$
where $$\Delta U$$ = change in internal energy, $$q$$ = heat exchanged, $$w$$ = work done.
For expansion/compression work against external pressure:
$$$w = -P_{\text{ext}} \Delta V = -P_{\text{ext}}(V_2 - V_1)$$$
So: $$\Delta U = q - P_{\text{ext}} \Delta V$$
Sign Convention
- Heat ($$q$$): Positive if absorbed by the system (endothermic); negative if released (exothermic)
- Work ($$w$$): Positive if done on the system; negative if done by the system
Worked Example: First Law
A gas absorbs 500 J of heat and expands, doing 200 J of work on the surroundings. Find $$\Delta U$$.
$$q = +500$$ J, $$w = -200$$ J (work done by system)
$$\Delta U = q + w = 500 + (-200) = \textbf{+300 J}$$
Work in Different Processes
Types of Thermodynamic Processes
- Isothermal (constant $$T$$): $$\Delta T = 0$$
- Isobaric (constant $$P$$): $$\Delta P = 0$$
- Isochoric (constant $$V$$): $$\Delta V = 0$$, so $$w = 0$$ and $$\Delta U = q_V$$
- Adiabatic (no heat exchange): $$q = 0$$, so $$\Delta U = w$$
Work Done by an Ideal Gas
Irreversible (against constant external pressure):
$$$w = -P_{\text{ext}}(V_2 - V_1)$$$
Reversible isothermal expansion/compression:
$$$w = -nRT \ln\frac{V_2}{V_1} = -2.303 \, nRT \log\frac{V_2}{V_1}$$$
Worked Example: Reversible Isothermal Work
2 mol of an ideal gas expands reversibly and isothermally at 300 K from 10 L to 20 L. Calculate the work done.
$$w = -nRT \ln\dfrac{V_2}{V_1} = -(2)(8.314)(300) \ln 2 = -4988.4 \times 0.693 = \textbf{-3457 J} \approx -3.46$$ kJ
Enthalpy
Enthalpy Change
$$$H = U + PV$$$
At constant pressure:
$$$\Delta H = \Delta U + P\Delta V \qquad \text{and} \qquad \Delta H = q_P$$$
For reactions involving ideal gases:
$$$\Delta H = \Delta U + \Delta n_g RT$$$
where $$\Delta n_g$$ = (moles of gaseous products) − (moles of gaseous reactants).
Worked Example: $$\Delta H$$ and $$\Delta U$$
For $$\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$$, if $$\Delta H = -92.4$$ kJ at 298 K, find $$\Delta U$$.
$$\Delta n_g = 2 - (1 + 3) = -2$$
$$\Delta U = \Delta H - \Delta n_g RT = -92.4 - (-2)(8.314 \times 10^{-3})(298) = -92.4 + 4.955 = \textbf{-87.4 kJ}$$
Exothermic vs Endothermic
| Property | Exothermic | Endothermic |
|---|---|---|
| Heat flow | Released to surroundings | Absorbed from surroundings |
| $$\Delta H$$ | Negative ($$< 0$$) | Positive ($$> 0$$) |
| Example | Combustion of fuels | Melting of ice |
Standard Enthalpy Changes
Using Formation Enthalpies
$$$\Delta_r H° = \sum \Delta_f H°(\text{products}) - \sum \Delta_f H°(\text{reactants})$$$
$$\Delta_f H°$$ of elements in their standard state = 0 (by definition).
Worked Example: Combustion of Methane
Calculate $$\Delta_r H°$$ for: $$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$
$$\Delta_r H° = [(-393.5) + 2(-285.8)] - [(-74.8) + 0] = -965.1 + 74.8 = \textbf{-890.3 kJ/mol}$$
Enthalpy from Bond Energies
$$$\Delta_r H° = \sum (\text{Bond energies of bonds broken}) - \sum (\text{Bond energies of bonds formed})$$$
Breaking bonds = positive (endothermic). Forming bonds = negative (exothermic).
Worked Example: Bond Energies
Estimate $$\Delta_r H°$$ for: $$\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)$$
Bond energies: H–H = 436 kJ/mol, Cl–Cl = 242 kJ/mol, H–Cl = 431 kJ/mol
Bonds broken: 436 + 242 = 678 kJ. Bonds formed: 2 × 431 = 862 kJ.
$$\Delta_r H° = 678 - 862 = \textbf{-184 kJ/mol}$$ (exothermic)
Hess's Law and Kirchhoff's Equation
Hess's Law
$$$\Delta H_{\text{reaction}} = \Delta H_1 + \Delta H_2 + \cdots$$$
Rules:
- If a reaction is reversed, the sign of $$\Delta H$$ changes.
- If a reaction is multiplied by a factor $$n$$, $$\Delta H$$ is also multiplied by $$n$$.
Kirchhoff's Equation
$$$\Delta H_2 = \Delta H_1 + \Delta C_P (T_2 - T_1)$$$
where $$\Delta C_P = \sum C_P(\text{products}) - \sum C_P(\text{reactants})$$
Entropy and the Second Law
Entropy Change
$$$\Delta S = \frac{q_{\text{rev}}}{T}$$$
For a reaction: $$\Delta_r S° = \sum S°(\text{products}) - \sum S°(\text{reactants})$$
Second Law of Thermodynamics
$$$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$$$
- $$\Delta S_{\text{universe}} > 0$$: Process is spontaneous
- $$\Delta S_{\text{universe}} = 0$$: Process is at equilibrium
- $$\Delta S_{\text{universe}} < 0$$: Process is non-spontaneous
Predicting Entropy Changes
Entropy increases ($$\Delta S > 0$$) when:
- Solid → liquid → gas (phase change increasing disorder)
- Number of gas moles increases ($$\Delta n_g > 0$$)
- Temperature increases
- A solute dissolves in a solvent
Gibbs Free Energy
Gibbs–Helmholtz Equation
$$$\Delta G = \Delta H - T\Delta S$$$
- $$\Delta G < 0$$: Reaction is spontaneous
- $$\Delta G = 0$$: Reaction is at equilibrium
- $$\Delta G > 0$$: Reaction is non-spontaneous
Effect of $$\Delta H$$ and $$\Delta S$$ on Spontaneity
| $$\Delta H$$ | $$\Delta S$$ | $$\Delta G$$ | Spontaneity |
|---|---|---|---|
| $$-$$ | $$+$$ | Always negative | Spontaneous at all $$T$$ |
| $$+$$ | $$-$$ | Always positive | Non-spontaneous at all $$T$$ |
| $$-$$ | $$-$$ | Negative at low $$T$$ | Spontaneous at low $$T$$ |
| $$+$$ | $$+$$ | Negative at high $$T$$ | Spontaneous at high $$T$$ |
Worked Example: Spontaneity
For a reaction, $$\Delta H = -10.5$$ kJ/mol and $$\Delta S = +31.4$$ J/(mol·K). Is it spontaneous at 298 K?
$$\Delta G = -10500 - (298)(31.4) = -10500 - 9357.2 = \textbf{-19857 J/mol} = -19.86$$ kJ/mol
Since $$\Delta G < 0$$, the reaction is spontaneous at 298 K.
Worked Example: Crossover Temperature
At what temperature does the decomposition of CaCO$$_3$$ become spontaneous?
Given: $$\Delta H = +178.3$$ kJ/mol, $$\Delta S = +160.4$$ J/(mol·K)
At threshold: $$\Delta G = 0$$, so $$T = \dfrac{\Delta H}{\Delta S} = \dfrac{178300}{160.4} = \textbf{1112 K} \approx 839°$$C
Tip: When $$\Delta H$$ and $$\Delta S$$ have the same sign, there is a crossover temperature $$T = \Delta H / \Delta S$$ where $$\Delta G = 0$$. JEE often asks you to find this temperature.
Gibbs Free Energy and Equilibrium
$$\Delta G$$ and Equilibrium Constant
$$$\Delta G = \Delta G° + RT \ln Q$$$
At equilibrium ($$\Delta G = 0$$, $$Q = K$$):
$$$\Delta G° = -RT \ln K = -2.303 \, RT \log K$$$
- $$\Delta G° < 0$$: $$K > 1$$ (products favoured)
- $$\Delta G° = 0$$: $$K = 1$$
- $$\Delta G° > 0$$: $$K < 1$$ (reactants favoured)
Worked Example: Equilibrium Constant from $$\Delta G°$$
Calculate $$K$$ at 298 K for a reaction with $$\Delta G° = -13.6$$ kJ/mol.
$$-13600 = -2.303 \times 8.314 \times 298 \times \log K$$
$$\log K = \dfrac{13600}{5705.8} = 2.383$$
$$K = 10^{2.383} \approx \textbf{241.5}$$
Summary of Key Formulas
Quick Reference
- First Law: $$\Delta U = q + w$$
- Enthalpy: $$\Delta H = \Delta U + \Delta n_g RT$$
- Reversible isothermal work: $$w = -nRT \ln(V_2/V_1)$$
- Hess's Law: $$\Delta H_{\text{rxn}} = \sum \Delta H_{\text{steps}}$$
- Kirchhoff: $$\Delta H_2 = \Delta H_1 + \Delta C_P(T_2 - T_1)$$
- Entropy: $$\Delta S = q_{\text{rev}} / T$$
- Gibbs: $$\Delta G = \Delta H - T\Delta S$$
- Equilibrium: $$\Delta G° = -RT \ln K$$
Important Note
The three most important equations for JEE from this chapter are: (1) $$\Delta U = q + w$$ (First Law), (2) $$\Delta H = \Delta U + \Delta n_g RT$$ (relating enthalpy and internal energy), and (3) $$\Delta G = \Delta H - T\Delta S$$ (Gibbs equation for spontaneity). Master these three and you can solve most thermodynamics problems.
Chemical Thermodynamics Formulas For JEE 2027: Conclusion
Preparing Chemical Thermodynamics well means understanding how internal energy, enthalpy, entropy, and Gibbs free energy are related. Focus on sign conventions, units, and the conditions under which each equation applies. Use a JEE Mains Chemistry Formula Sheet to revise key results, then practise questions on heat changes, Hess’s law, spontaneity, and equilibrium to strengthen your understanding.
Refer to JEE Mains Chemistry chapter-wise weightage when planning your revision, while remembering that question distribution can vary across papers. Combine formula revision with previous-year questions and timed practice, and review mistakes carefully. This approach can help you improve calculation accuracy and tackle thermodynamics problems with greater confidence.
